Answer
The orbital body and comet can intersect at $\left( 0,-4 \right),\left( -2,0 \right),\left( 2,0 \right)$
Work Step by Step
Let us consider the paths followed by the orbital body and the comet
$\begin{align}
& 16{{x}^{2}}+4{{y}^{2}}=64\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& y={{x}^{2}}-4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right)
\end{align}$
Put the value of $ y={{x}^{2}}-4$ into equation (I):
$\begin{align}
& 16{{x}^{2}}+4{{\left( {{x}^{2}}-4 \right)}^{2}}=64 \\
& 16{{x}^{2}}+4\left( {{x}^{4}}-8{{x}^{2}}+16 \right)=64 \\
& 16{{x}^{2}}+4{{x}^{4}}-32{{x}^{2}}+64=64 \\
& 4{{x}^{4}}-16{{x}^{2}}=0
\end{align}$
Further solving it,
$\begin{align}
& {{x}^{4}}-4{{x}^{2}}=0 \\
& {{x}^{2}}\left( {{x}^{2}}-4 \right)=0 \\
& {{x}^{2}}=0 \\
& x=0
\end{align}$
Substitute the value of $ x=0,2,-2$ in equation (I):
If,
$\begin{align}
& x=0 \\
& y={{\left( 0 \right)}^{2}}-4 \\
& =-4
\end{align}$
If,
$\begin{align}
& x=2 \\
& y={{\left( 2 \right)}^{2}}-4 \\
& =0
\end{align}$
If,
$\begin{align}
& x=-2 \\
& y={{\left( -2 \right)}^{2}}-4 \\
& =0
\end{align}$
Thus, the possible points of intersection are $\left( 0,-4 \right),\left( -2,0 \right),\left( 2,0 \right)$.
